avicenna

15 hours ago, Carrock said:

There's been a bit of conflation in some replies between A.C. and D.C..  D.C. is conserved but A.C. isn't.

So: V.H.F. A.C. whose effects are more obvious than 50Hz.

You could have 100 meters of 50 ohm coaxial cable with 10db loss per 100 meters. Connecting A.C. power at say 100MHz 100V to the (resistive) cable gives I =V/R = 2 amps input (200W). Connect a 50 ohm resistive load to the far end and you'll get ~ 31.6V at 0.632A i.e. 20W output. If you use a 200 meter cable you'll get 10V at 0.2A i.e. 2W. The input is still 200W into 50 ohms.

The current drops exponentially along the cable. The main power losses are I^R losses in the conductor, dielectric (insulator) heating and radiation from the cable. There is no A.C. current conservation; some of it charges and discharges the dielectric and current is also involved in creating magnetic and electromagnetic fields.

One way of dealing with reactance is to consider the effect of a load impedance mismatch.

e.g. terminate the cable with 25 instead of 50 ohms. This will cause a power reflection back into the cable to compensate for trying to connect a 50 ohm cable to a 25 ohm load. If 20W output then 20 *(50 -25)/(50+25)W i.e. 6.7W is reflected back into the cable and after attenuation 0.67W reaches the source.

You'll get standing waves on the cable; every half wavelength (About 1.5m) you'll get maximum voltage and minimum current; between these nodes you get a minimum voltage, maximum current node. A.C. current can be created and destroyed without breaking conservation laws. Some energy is stored in various fields and doesn't reach the load; sometimes it's called imaginary power(it can be treated as 90 deg or sqrt(-1) out of phase with 'real' power) or reactive power(capacitors and inductors have reactance).

This is sort of real; there are meters which measure forward and reflected power in coax cables...

I didn't want to oversimplify too much; this post ended up much longer than I intended.

I believe this is the real deal, but only for those with some foundations in AC circuit theory. I will see how things relate to what I am investigating.  

@Carrock. 

I am trying to reduce AC circuit to DC circuit "instantaneously". 

I don't want coaxial cables which complicates things. Say I have a simple ac generator that generates fairly good sinusoidal voltage source at 50 Hz (If possible at all?). I connect a long resistive wire to the terminals in a huge circular loop. When the wire is at thermal equilibrium with the environment, we know that all power will be dissipated as IR radiation loss, purely resistive loss - assume ideally.

So we could always apply ohm's law of I=V/R where R is the resistance of the wire, V the instantaneous voltage. It seems that there will be the usual charge conservation along the wire as if the current is a dc current. The current should be a constant at that moment of consideration.

My setup would eliminate capacitance, inductance etc. Instantaneously, we only have the magnetic fields around the wire which we assume "steady". How is such an analysis.
 

2 minutes ago, swansont said:

The latter part of my explanation applies to this. The current is uniform/\. There’s no way for it to vary.

Thanks. This is what I want to confirm.

For the AC case, I really cannot say as the real world and the ideal world may be different. 

3 minutes ago, swansont said:

If the voltage is constant (for a real wire) there would be no current. If there is a voltage drop, and thus a current, the current will be uniform even if the voltage drop is not (e.g. if there’s a resistor, or a series of different-valued resistors); there’s no way to vary it. Charge is conserved, so current flowing in to a point equals the current flowing out.

I am saying the potential difference between the ends of the wire, i.e. a long wire connected to a dc battery.

36 minutes ago, sethoflagos said:

I think the correct answer to this rather odd question is yes, though if the wire has any resistance then that uniform current is zero. In the absence of reactive elements (capacitors or inductors) a constant current implies a constant voltage gradient.

I don't  understand. I = V/R, so how can there be zero current if we connect a AA 1.5 V battery to a long copper wire. My question is whether I is the same at all points of the wire.  

If the voltage across a long wire is constant, is the current uniform throughout the wire length.  

In a 50 Hz ac voltage across a long wire, at a certain moment how does current vary along x, the wire length. 
 

Maybe no easy answer because we know too little yet about light.

The wiki says it is the bound electrons that absorbs radiation; then how the electrons energy get transferred  to the nucleus kinetic energy. I think temperature not dependent on the KE of electrons, but only in the KE of the nucleus or center of mass.  

I'll like to know the actual physical mechanism how matter absorbs EM radiation. Take the specific example how copper metal absorbs IR radiation and the copper having its temperature raised.
 

In a capacitor discharge through a plain resistor, the capacitor power supplied at any instant is VI; the power dissipated in the resistor is I²R. So VI = I²R.

Consider a railgun operated with a capacitor bank. At any instant of capacitor discharge, the power supplied is VI. The total power supplied for ohmic loss is sum I²R for two rails plus the resistance of the armature.

Question: Since VI = total I²R, how can the power equation include the kinetic energy supplied to the armature?

I think correct about plasma.

7 hours ago, Sensei said:

Any object with temperature T is emitting photons. How much energy have these photons depend on temperature of the body. Your biological body is also emitting infra-red photons (therefor one can use IR thermometer to remotely check temperature, or IR camera to detect people or animals from longer distance). If body is enough hot, there are emitted visible photons (and your eyes are able to detect them as light).

Please read black-body radiation article on Wikipedia for more details about this subject:

https://en.wikipedia.org/wiki/Black-body_radiation

I think this is the answer. Thanks.

"he continuous emission spectrum is due to the surface temperature of the Sun.

2 minutes ago, studiot said:

The continuous emission spectrum is due to the surface temperature of the Sun.

This is obviously hotter than that of a steel bar heated to red or white heat, but the principle is the same.

 

According to the Bohr model, light of a specific frequency is emitted due to the difference of two energy states of one (? or a few) element. Why a white hot iron bar can emit white light?  

The spectrum of sunlight is a continuous spectrum as in the rainbow. But superimposed on the spectrum are dark absorption lines of some specific frequencies. From the absorption lines, we are able to tell that the main elements of the sun is hydrogen, iron, carbon, helium and some others.

My question is why the continuous spectrum ? Elements have characteristic emission line spectra. So every specific wavelength in the sun's continuous spectrum is associated with one (?) element which has that wavelength in its line spectrum. Does it mean that the sun has all 108+ known elements?      

[edit] Or the emission lines of the main elements in the sun - iron,carbon, etc. - sufficient to form the continuous spectrum.

Yes.This Steiner paper exactly satisfy what I am looking for. It gave the first value of h to around end 1910. It also gave the method used to measure h historically, basically the photoelectric effect in the earlier years. I am not able to access any peer reviewed articles. Thanks.

I posted the same question over physics stack exchange. Because they can't give me the information, they insist that there is no reason for me to to look for such information as the "different values should not be comparable". At other time physics stackexchange do give good answers.

It would probably mean you and I are not up-to-mark to view those papers. 

I looked all over to get the various accepted value of the Planck constant since 1900. But there is never
any record about the history of this mysterious number.  

36 minutes ago, Strange said:

This just as direct as Bragg!s law

You may say so. But if Bragg's method can be used for all gamma ray, does it not bypass the relation E=hν. What if there is a method so "direct" that it requires 99 more relation in between?

Can Bragg's method be used for all gamma rays?

5 hours ago, studiot said:

 

5 hours ago, studiot said:

Why is this not direct?

What do you mean by direct?

Frequency analysis of gamma rays is called gamma ray spectroscopy

https://en.wikipedia.org/wiki/Gamma_spectroscopy

Note gamma and X ray frequencie ranges sunstantially overlap.

The difference is the nature of source.

Bragg's law : nλ = 2d Sinθ relates wavelength of x-rays "directly" to angle. I consider this direct.

Can this be used for all gamma rays.

In E=hf, we need to assume an energy to frequency relation and so it is "indirect."

4 hours ago, studiot said:

Why is this not direct?

What do you mean by direct?

Frequency analysis of gamma rays is called gamma ray spectroscopy

https://en.wikipedia.org/wiki/Gamma_spectroscopy

Note gamma and X ray frequencie ranges sunstantially overlap.

The difference is the nature of source.

Then, I will re-phrase. Measuring gamma ray frequencies without using energy detector and E=hf.  

1 hour ago, swansont said:

You can do diffraction with x-rays and gammas. Crystal lattices have spacing that is appropriate for this. X-ray crystallography would work with some gammas

1 hour ago, swansont said:

You can do diffraction with x-rays and gammas. Crystal lattices have spacing that is appropriate for this. X-ray crystallography would work with some gammas

Then what about those outside of the possible "some" ranges; the higher known frequencies.

I know there are ways to measure light frequencies directly using diffraction, etc, even for x-rays.

Is there a way to directly measure the frequency of gamma rays? Not using energy detectors and E=hf.

39 minutes ago, Strange said:

No.

Although, if they have the same frequency, then you could get constructive or destructive interference: https://en.wikipedia.org/wiki/Wave_interference

Isn't it strange that light-light do not interact?

Say if I have a laser beam. If I shine another laser beam to intersect the first beam,
will this first beam be affected?

1 hour ago, John Cuthber said:

They plan to.
https://www.nist.gov/si-redefinition/kilogram-silicon-spheres-and-international-avogadro-project

This is only about a redefinition of some SI units such as the kilogram. With whatever changes in the SI units, the atomic mass will still be measured through mass spectrometry.

When mass spectrometry was invented after the 1910, they only go for greater improvements to get higher resolution. There is no published experiment that anyone made any independent verification if mass spectrometry is reliable. Our current analytical balance is accurate to 10¯⁵, enough to make chemical analysis of relative atomic mass of compounds composed of single isotopes of the elements. A simple example is sodium iodide; both elements exist as stable single isotope in nature.        

Has there been any experiment done through chemical analysis to check if mass spectrometry is consistent with our chemical balance?

Our current analytical balance has an accuracy of 10¯⁵. The chemical composition by weight of two isotopes forming a compound could be analyzed. This is sufficient to determine a relative atomic mass of two isotopes and to compare with the CODATA values obtained through mass spectrometry. The values should be the same to at least the third decimal. This is a very good test of the reliability of mass spectrometry.  

We have about 19 monoisotopic elements including Fluorine and Iodine; these two may react with the other monoisotopic elements to form compounds.

14 minutes ago, John Cuthber said:

The atomic masses were determined (often to quite high accuracy) well before there were any mass spectrometers.

For example, you can take a gram of hydrogen(mixed with nitrogen r something else that won't react), and pass it over hot copper oxide.

The hydrogen will be oxidised to water.
You can trap that water by passing the gas over phosphorus pentoxide.

And you can weigh the water.

You will get 9 grams of water and so you know that the mass of oxygen that combines with 1 gram of hydrogen is 8 grams .

It gets a bit more complicated when you try to decide if water is HO H₂O HO₂ or what, but it was pretty much all worked out by classical chemistry.

 

What you get is only the classical atomic weights. In nature H has 0.02% deuterium. O has O16, O17 and O18 in various %. So the old atomic weights is a mixture of isotopes. We need H₂O that is made from pure isotope ¹H and pure isotope ¹⁶O. Only then can we get real relative atomic mass to compare with the CODATA obtained from mass spectrometry..

36 minutes ago, John Cuthber said:

Or with a questionable definition of "chemistry"...

https://en.wikipedia.org/wiki/Mass_spectrometry

 

It is possible to observe isotope mass effects with chemistry.

According to this data
https://en.wikipedia.org/wiki/Isotopes_of_lead

you can get samples of lead that have between 51 and 56 % 208Pb

That would be a big enough range to detect if you measured the molar mass by classical wet chemistry.

I know isotopes exists. What I am really interested is getting the actual relative atomic mass using our chemical balances and not through mass spectrometry. Then compare it with the CODATA.

Has anyone ever checked if their scales don't give short weights and measures! Even in ancient times, officials would come to check the scales you use to measure grains. What if mass spectrometry is off! Avicenna would even ask if God exists.

I know we can get the relative atomic mass of any two nuclides by just looking up the CODATA.
But I'll like to know if anyone has measured atomic mass with a chemical process.

Say we analyze the weight composition of ¹H₂¹⁶O to get the relative mass of ¹⁶O/¹H. Or of any other two elements through any chemical process.